Given that the inverse trigonometric function assumes principal values only. Let x, y be any two real numbers in `[-1, 1]` such that `cos^(-1)x – sin^(-1)y = alpha, frac{-pi}{2} le alpha le pi`.
Then, the minimum value of `x^2 + y^2 + 2xy sin alpha` is

Options

A. `-1`
B. `1/2`
C. `(-1)/2`
D. 0

Answer

To solve the problem, we need to find the minimum value of the expression \( x^2 + y^2 + 2xy \sin \alpha \) given that \( \cos^{-1} x - \sin^{-1} y = \alpha \) where \( x, y \in [-1, 1] \) and \( -\frac{\pi}{2} \leq \alpha \leq \pi \).

### Step-by-Step Solution:

1. **Rewrite the equation**:
We start with the equation:
\[ \cos^{-1} x - \sin^{-1} y = \alpha \]
We can rewrite \( \sin^{-1} y \) as \( \frac{\pi}{2} - \cos^{-1} y \):
\[ \cos^{-1} x - \left(\frac{\pi}{2} - \cos^{-1} y\right) = \alpha \]
Simplifying this gives:
\[ \cos^{-1} x + \cos^{-1} y = \alpha + \frac{\pi}{2} \]

2. **Use the cosine addition formula**:
Using the property of cosine, we have:
\[ \cos(\cos^{-1} x + \cos^{-1} y) = -\sin\left(\alpha + \frac{\pi}{2}\right) \]
Since \( \sin\left(\alpha + \frac{\pi}{2}\right) = \cos \alpha \), we can write:
\[ xy - \sqrt{(1-x^2)(1-y^2)} = -\cos \alpha \]

3. **Rearranging the equation**:
Rearranging gives:
\[ xy + \cos \alpha = \sqrt{(1-x^2)(1-y^2)} \]

4. **Square both sides**:
Squaring both sides yields:
\[ (xy + \cos \alpha)^2 = (1-x^2)(1-y^2) \]
Expanding both sides gives:
\[ x^2y^2 + 2xy \cos \alpha + \cos^2 \alpha = 1 - x^2 - y^2 + x^2y^2 \]
Simplifying this leads to:
\[ 2xy \cos \alpha + \cos^2 \alpha = 1 - x^2 - y^2 \]

5. **Rearranging the expression**:
Rearranging gives:
\[ x^2 + y^2 + 2xy \cos \alpha = 1 - \cos^2 \alpha \]
Since \( 1 - \cos^2 \alpha = \sin^2 \alpha \), we have:
\[ x^2 + y^2 + 2xy \sin \alpha = \sin^2 \alpha \]

6. **Finding the minimum value**:
The expression \( x^2 + y^2 + 2xy \sin \alpha \) can be minimized. The minimum value of \( \sin^2 \alpha \) occurs when \( \alpha = 0 \) (since \( \sin \alpha \) ranges from -1 to 1). Thus, the minimum value of \( \sin^2 \alpha \) is 0.

### Conclusion:
Therefore, the minimum value of \( x^2 + y^2 + 2xy \sin \alpha \) is:
\[ \boxed{0} \]